Question

If one starts with pure NO2(g) at a pressure of 0.500 atm, the total pressure inside the reaction vessel when 2NO2(g) 2NO(g) + O2(g) reaches equilibrium is 0.674 atm. Calculate the equilibrium partial pressure of NO2.

Answer 1

Answer:

The partial pressure of NO2  = 0.152 atm

Explanation:

Step 1: Data given

Pressure NO2 = 0.500 atm

Total pressure at equilibrium = 0.674 atm

Step 2: The balanced equation

2NO2(g) → 2NO(g) + O2(g)

Step 3: The initial pressure

pNO2 = 0.500 atm

pNO = 0 atm

p O2 = 0 atm

Step 4: Calculate pressure at the equilibrium

For 2 moles NO2 we'll have 2 moles NO and 1 mol O2

pNO2 = 0.500 - 2x atm

pNO =2x atm

pO2 = xatm

The total pressure = p(total) = p(NO2) + p(NO) + p(O2)

p(total) = (0.500 - 2x) + 2x + x= 0.674 atm

0.500 + x = 0.674 atm

x = 0.174 atm

This means the partial pressure of NO2 = 0.500 - 2*0.174 = 0.152 atm

Answer 2

The partial pressure of $\rm NO_2$ at equilibrium has been 0.152 atm.

The partial pressure has been the pressure that has been exerted by the single gas in the mixture. The given:

$\rm NO_2$ pressure = 0.5 atm.

The total pressure = 0.674 atm.

At the equilibrium, the reaction pressure has been given as:

$\rm 2\;NO_2\;\rightarrow\;2\;NO\;+\;O_2$

0.5             0            0      I

-2x             2x         x       C

0.5 - 2x     +2x       x        E

According to the law of conservation:

Total pressure = sum of the partial pressure of each gas:

0.674 atm = 0.5 - 2x + 2x + x

0.674 atm = 0.5 + x

x = 0.174 atm.

The partial pressure of $\rm NO_2$ at equilibrium has been 0.5 - 2x

= 0.5 - 2 × 0.174 atm.

= 0.152 atm

The partial pressure of $\rm NO_2$ at equilibrium has been 0.152 atm.

For more information about the partial pressure, refer to the link:

https://brainly.com/question/14623719